Note:
For the benefit of the students, specially the aspiring ones, the question of JEE(advanced), 2020 are also
given in this booklet. Keeping the interest of students studying in class XI, the questions based on topics
from class XI have been marked with ‘*’, which can be attempted as a test. For this test the time
allocated in Physics, Chemistry & Mathematics are 20 minutes, 30 minutes and 25 minutes respectively.
FIITJEE
SOLUTIONS TO JEE (ADVANCED) – 2020
PA R T I: PH YSI CS
Section - 1 (Maximum Marks: 18)




*1.
This section contains SIX (06) questions.
Each question have FOUR options. ONLY ONE of these four options is the correct answer.
For each question, choose the option corresponding to the correct answer.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks
: +3 If ONLY the correct option is chosen.
Zero Marks
: 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks
: –1 In all other cases.
A football of radius R is kept on a hole of radius r (r < R) made on a plank kept horizontally. One end of
the plank is now lifted so that it gets tilted making an angle  from the horizontal as shown in the figure
below. The maximum value of  so that the football does not start rolling down the plank satisfies (figure is
schematic and not drawn to scale)
r
R
r
(C) sin  
2R
(A) sin  
Sol.
r
R
r
(D) cos  
2R
(B) tan  
A
At the verge of toppling, N1 = 0
h
mg sinh = mg cos  r and cos  
R
h
So, mgsin h  mg r
R
 sin  = r/R
N
Plank
N1
R
2r
mg cos

/2
mg sin 
mg

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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2
2.
A light disc made of aluminium (a nonmagnetic material) is kept horizontally and is free to rotate about its
axis as shown in the figure. A strong magnet is held vertically at a point above the disc away from its axis.
On revolving the magnet about the axis of the disc, the disc will (figure is schematic and not drawn to
scale)
(A) rotate in the direction opposite to the direction of magnet’s motion
(B) rotate in the same direction as the direction of magnet’s motion
(C) not rotate and its temperature will remain unchanged
(D) not rotate but its temperature will slowly rise
Sol.
*3.
Sol.
B
 is decreasing, so clockwise current in the square loop is induced.
So force (along the arrow) is acting on the disc.
So movement of disc is along the direction of motion of magnet.

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a
horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of
the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without
slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the
position of the scale will look like (figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
B
v  R = 0
v
r

r  v 1  
R
 R
5 

So, xP  50  1 
  75 cm
 10 
P
vp  v  r  v 
R
r
v

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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3
4.
A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two
ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are
connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field
Bo parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the
coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum
gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during
this time)

NQB0 R 2
2
(C) 2NQB0 R 2
(B) NQB0 R 2
(A)
(D) 4NQB0 R 2
Sol.
B
A.M.   dt   NIR2Bdt  NR 2B0 Q
5.
A parallel beam of light strikes a piece of transparent glass having cross section as shown in the figure
below. Correct shape of the emergent wavefront will be (figures are schematic and not drawn to scale)
Sol.
(A)
(B)
(C)
(D)
A
 Emerging wave front
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-4
*6
An open-ended U-tube of uniform cross-sectional area contains water (density 103kg m–3). Initially the
water level stands at 0.29 m from the bottom in each arm. Kerosene oil (a water-immiscible liquid) of
density 800 kg m–3 is added to the left arm until its length is 0.1 m, as shown in the schematic figure below.
h 
The ratio  1  of the heights of the liquid in the two arms is
 h2 
15
14
7
(C)
6
35
33
5
(D)
4
(A)
Sol.
(B)
B
800  0.1 + 1000  x = 1000  (0.58  x)
2000 x = 500
x = 0.25 m
So, h1 = 0.35 m, h2 = 0.33 m
h1 35

h2 33
x
h2 = 0.58
x
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-5
Section 2 (Maximum Marks: 24)




7.
This section contains SIX (06) questions.
Each question have FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) the
correct answer(s).
For each question, choose the option(s) corresponding to (all) the correct answer(s).
Answer to each question will be evaluated according to the following marking scheme:
Full Marks
: +4 If only (all) the correct option(s) is(are) chosen.
Partial Marks : +3 all the four the correct option but ONLY three options are chosen.
Partial Marks : +2 If three or more options are correct but ONLY two are chosen both of which are
correct.
Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and it is a
correct option.
Zero Marks
: 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : –2 In all other cases.
A particle of mass m moves in circular orbits with potential energy V(r) = Fr, where F is a positive constant
and r is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the
particle’s orbit is denoted by R and its speed and energy are denoted by v and E, respectively, then for the
nth orbit (here h is the Planck’s constant)
(A) R  n1/3 and v  n 2/3
(B) R  n 2/3 and v  n1/3
1/3
3  n 2 h 2 F2 
(C) E   2 
2  4 m 
Sol.
1/3
 n 2 h 2 F2 

2
 4 m 
(D) E  2 
B, C
F
dV
dr
mv 2
F
R
nh
mvr 
2
Solving above equations
n2h2
R3  2
4 m
v  n1/3
8.
The filament of a light bulb has surface area 64 mm2. The filament can be considered as a black body at
temperature 2500 K emitting radiation like a point source when viewed from far. At night the light bulb is
observed from a distance of 100 m. Assume the pupil of the eyes of the observer to be circular with radius
3 mm. Then
(Take Stefan-Boltzmann constant = 5.67×10–8Wm–2K–4, Wien’s displacement constant
=2.90×10–3 m-K, Planck’s constant = 6.63×10–34 Js, speed of light in vacuum = 3.00×108 ms–1)
(A) power radiated by the filament is in the range 642 W to 645 W
(B) radiated power entering into one eye of the observer is in the range 3.15×10–8 W to 3.25×10–8 W
(C) the wavelength corresponding to the maximum intensity of light is 1160 nm
(D) taking the average wavelength of emitted radiation to be 1740 nm, the total number of photons
entering per second into one eye of the observer is in the range 2.75×1011 to 2.85×1011
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-6
Sol.
B, C, D
P = AT4
= 64  106  5.67  10 8  (2500)4  140W
Power entering in eye
140
140  9  10 10
3 2
=

(3

10
)

 3.15  108 W
4(100)2
4
T  b   
2.93  106
   1160
2500
r=
3mm
100m
Photons
12420 3.15  108
n

 2.75  1011
1740
1.6  10 19
9.
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of
only one physical quantity. In one such system, dimensions of different quantities are given in terms of a
quantity X as follows: [position] = [X]; [Speed] = [X]; [acceleration] = [Xp]; [Linear momentum] = [Xq];
[force] = [Xr]. Then
(A)   p  2
(B) p  q  r  
(C) p  q  r  
Sol.
(D) p  q  r  
A, B
r  x
v  x
a  xP
P  xq
F  xr
(A) ra   x P 
 Energy 
 mass   [2]


 aP 
(B)    at    v   
F 
P+qr=
10.

A uniform electric field, E  400 3yˆ NC 1 is applied in a region. A charged particle of mass m carrying
positive charge q is projected in this region with an initial speed of 2 10 106 ms1. This particle is aimed
to hit a target T, which is 5 m away from its entry point into the field as shown schematically in the figure.
q
Take
 1010 Ckg 1. Then
m
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-7
(A) the particle will hit T if projected at an angle 45 from the horizontal
(B) the particle will hit T if projected either at an angle 30 or 60 from the horizontal
5
5
(C) time taken by the particle to hit T could be
μs as well as
μs
6
2
(D) time taken by the particle to hit T is
Sol.
5
μs
3
B, C
Range = R =
2usin 
ucos   5m
geff
geff =
(qE/m)
3
 qER 
sin2  

2 
mu
2


 = 30
So for same range angle of projection is either 30 or 60
So Option (B) is correct

2u sin   10
T

 10 6  sin 
 3

geff


T1 (  30) 
5
s
6
5
s
2
So, Option (C) is correct
T2 (  60) 
11
Shown in the figure is a semicircular metallic strip that has thickness t and resistivity  . Its inner radius is
R1 and outer radius is R2. If a voltage V0 is applied between its two ends, a current I flows in it. In addition,
it is observed that a transverse voltage V develops between its inner and outer surfaces due to purely
kinetic effects of moving electrons (ignore any role of the magnetic field due to the current). Then (figure is
schematic and not drawn to scale)
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-8
V
R2
R1
I
(A) I 
I
V0
v0 t  R 2 
ln 

  R1 
(B) the outer surface is at a higher voltage than the inner surface
(C) the outer surface is at a lower voltage than the inner surface
(D) V  I 2
Sol.
A, C, D
Area of the strip = (t.dr)
 r 
dR   

 tdr 
dr
r
R
2
1
tdr
 dR   r
R1
 t  R 
1
   Ln  2 
R eq     R1 
I
I
V0

Req 
R 
t  n  2 
 R1 
So, I 
V0  V0 t   R2 

 So, option (A) is correct.
 n 
Req     R1 
ET
Electrons are revolving in a circular path so there is a centripetal force
on the electron, which provide a transverse electric field.
The direction of transverse electric field is radially out side hence
inner surface is at higher potential.
So, option (C) is correct.
Vd
I
I
V0
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-9
Current density J(r) 
dI
dI
(tdr)
V0
(dr)(tdr)
V0
J(r) 
 r 

 (tdr)
 tdr 
V0
J(r) 
(r)
r
J(r) 
I = neAVd
J(r) = neVd
 J(r)   V0 
 Vd  


 ne   rne 
Centripetal force
m v2
eET  e d
r
From (i) and (ii) ET  V02
I
I
V0
…(i)
…(ii)
….(iii)
V   E T dr
 V  ET
….(iv)
From (iii) and (iv)
V  V02
I  V0
Hence
V  I2
Hence Option (D) is correct.
*12
As shown schematically in the figure, two vessels contain water solutions (at temperature T) of potassium
permanganate (KMnO4) of different concentrations n 1 and n2 (n1>n2) molecules per unit volume with
Δn=(n1−n2)<<n1. When they are connected by a tube of small length  and cross-sectional area S, KMnO4
starts to diffuse from the left to the right vessel through the tube. Consider the collection of molecules to
behave as dilute ideal gases and the difference in their partial pressure in the two vessels causing the
diffusion. The speed v of the molecules is limited by the viscous force −v on each molecule, where  is a
constant. Neglecting all terms of the order (Δn)2, which of the following is/are correct? ( kB is the
Boltzmann constant)
(A) the force causing the molecules to move across the tube is  nkBTS
(B) force balance implies n1 vl =  nkBT
 n   k B t 
(C) total number of molecules going across the tube per sec is 
S

 l   
(D) rate of molecules getting transferred through the tube does not change with time
Sol.
A, B, C
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 0
PV = NKT
P1 = n1KBT
P2 = n2KBT
n1, n2  no. of molecules per unit volume
Force acting on the molecules
F = PS
=(P1  P2)S
F = (n1  n2)KBTS = nkBTS…(i)
So Option (A) is correct
Total no. of molecules in the tube = (n1S)
Force acting on each molecules
F = v
So total force acting on (n1s) molecules = 

P1
P2
n1
S

(v)(n1s)…(ii)
From (i) and (ii)
nKBTS = (v)(n1S)
So, nKBT = (vn1)…(iii)
So, Option (B) is correct.
No. of molecules going across the tube per second
= (svn1)
 nK  T   SnK T 


 n1    
= (sn1 ) 
So, Option (C) is correct.
With respect to time n changes hence rate of
molecules getting transferred through the tube
changes with time.
So, Option (D) is incorrect.
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 1
Section 3 (Maximum Marks: 24)



This section contains SIX (06) questions. The answer to each question is a NUMERICAL VALUE
For each question, enter the correct numerical value of the answer using the mouse and on-screen virtual
numerical keypad in the place designated to enter the answer. If the numerical value has more than tow
decimal places, truncate/round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following marking scheme:
Full Marks
: +4 If ONLY the correct numerical value is entered;
Zero Marks
: 0 If all the cases.
*13.
Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the
right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only
the left finger slips with respect to the scale and the right finger does not. After some distance, the left
finger stops and the right one starts slipping. Then the right finger stops at a distance xR from the center
(50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the
frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers
and the scale are 0.40 and 0.32, respectively, the value of xR (in cm) is ______.
Sol.
25.6 (25.60 to 25.60)
Initially
40N1  50N2
CM
N2
For first move
kN3   sN4
xN3  40 N4
x = 32 cm
50 cm
1
CM
…(i)
…(ii)
For second move
 sN5  kN6
…(i)
32(N5 )  XR N6 …(ii)
XR = 25.6 cm
40 cmN
N3
x cm
40 cmN
4
CM
32 cm
N5
xR N
6
*14.
When water is filled carefully in a glass, one can fill it to a height h above the rim of the glass due to the
surface tension of water. To calculate h just before water starts flowing, model the shape of the water above
the rim as a disc of thickness h having semicircular edges, as shown schematically in the figure. When the
pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension, the
water surface breaks near the rim and water starts flowing from there. If the density of water, its surface
tension and the acceleration due to gravity are 103kg m−3, 0.07 Nm−1 and 10 ms−2, respectively, the value of
h (in mm) is _________.
Sol.
3.75 (3.65 to 3.85)
 gh 
S(2R)  
 h(2R)
 2 
h = 3.75  103 mm
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 2
15.
One end of a spring of negligible unstretched length and spring constant k is fixed at the origin (0,0).
A point particle of mass m carrying a positive charge q is attached at its other end. The entire system is kept

on a smooth horizontal surface. When a point dipole p pointing towards the charge q is fixed at the origin,
the spring gets stretched to a length l and attains a new equilibrium position (see figure below). If the point
mass is now displaced slightly by  l << l from its equilibrium position and released, it is found to oscillate
at frequency
Sol.
1 k
. The value of ߜ is ______.
 m
3.14 (0.50 to 0.50) or (3.13 to 3.15)
Potential energy of a system after a small displacement.
1
k pq
U  kx 2 
2
(  x)2
dU
2k pq
 kx 
dx
(  x)3
d2U
6k pq
k
 k  3k  4k
2
dx
(  x)4
f
1 4k
2 m
1 k
 m
So,  = 3.14
and for angular frequency
f
1 k
 m
 = 0.5

*16.
Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It expands
isothermally to volume 4V1. After this, the gas expands adiabatically and its volume becomes 32 V1. The
work done by the gas during isothermal and adiabatic expansion processes are Wiso and Wadia, respectively.
If the ratio
Sol.
Wiso
=݂ ln2, then ݂ is ________.
Wadia
1.77 (1.77 to 1.79)
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 3
AB
P1V1 = PB  4V1
P
 PB  1
4
Wiso  1 R  Tn
P1
4V1
V1
B
P1/4
P1V1n4  2P1V1n2
BC
PB VB  PC VC
C
V1
P1
(4V1 )  PC (32V1 )
4

A
4V1
32V1
5 /3
P1
1 1
1 1 
  P1   3  P1 =
48
42 
128
P1
P1
 32V1   4V1
4
Wadic  128
5
1
3
P1V1
3
 P1V1  P1V1
9
= 4
 4
 P1V1
2
2
8


3
3
Wiso
2P V n2 16
 1 1

n2
9
Wadio
9
P1V1
8
Comparing with given expression of = 16/9 = 1.78
 PC 
*17.
A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a
speed of 2 ms−1 in front of the open end of the pipe and parallel to it, the length of the pipe should be
changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is 320 ms−1,
the smallest value of the percentage change required in the length of the pipe is ____________.
Sol.
0.62 (0.62 to 0.64)
320
320
f  f0

f0
320  2 318
2
 f 
f0
318
(2n  1)v
 f0 
4L
2
L

318
L
L 200
100 

 0.62983  0.63
L
318
…(1)
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 4
18.
Sol.
 r
A circular disc of radius R carries surface charge density  0  r    0 1   , where 0 is a constant and r
 R
is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the
R
charged disc completely is 0. Electric flux through another spherical surface of radius
and concentric
4

with the disc is  . Then the ratio 0 is_________

6.40 (6.40 to 6.40)
Total charge on disc
R
R
R

r
1

q0   0  1   2rdr    2   rdr   r 2 dr 
R0
 R
0
0

R2 R2 

= 2 

3 
 2
 R2 
q0  2  
 6 
Change enclosed by second sphere
R/ 4
R/ 4

1
q0    2   rdr   r 2 dr 
R 0
0

2
2
R
R 

= 2 

32
192


…(1)
 5R2 
q0  2 

 192 

q
R2 / 6
192 32
 0  0 


 6.40
2

q 5R / 192 5  6
5
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 5
PA R T II: C H E MIST RY




*1.
SECTION-1 (Maximum Marks: 18)
This section contains SIX (06) questions
Each question has FOUR options. ONLY ONE of these four options is the correct answer.
For each question, choose the option corresponding to the correct answer.
Answer to each question will be evaluated according to the following making scheme:
Full Marks
: +3 If ONLY the correct option is chosen;
Zero Marks
: 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : 1 In all other cases
If the distribution of molecular speeds of a gas is as per the figure shown below, then the ratio of the most
probable, the average, and the root mean square speeds, respectively, is
(A) 1:1:1
(C) 1 : 1.128 : 1.224
Sol.
(B) 1 : 1 : 1.224
(D) 1 : 1.128 : 1
B
Since the graph is symmetrical about the peak. So, the average speed and most probable speed will coincide
but rms speed will remain
So, ump 
urms 
3RT
M
2RT
 uav
M
3RT
M
ump : uav : urms  2 : 2 : 3  1: 1:
3
2
 1: 1: 1.224
*2.
Which of the following liberates O 2 upon hydrolysis?
(A) Pb3 O4
(C) Na2 O2
Sol.
(B) KO 2
(D) Li2O2
B
Pb3 O4  H2 O  No reaction
KO2  H2O 
 KOH  H2 O2  O2
Na2 O2  H2O  NaOH  H2 O2
Li2O2  H2O LiOH  H2 O2
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 6
3.
A colorless aqueous solution contains nitrates of two metals, X and Y. When it was added to an aqueous
solution of NaCl, a white precipitate was formed. This precipitate was found to be partly soluble in hot
water to give a residue P and a solution Q. The residue P was soluble in aqueous NH3 and also in excess
sodium thiosulfate. The hot solution Q gave a yellow precipitate with KI. The metals X and Y, respectively
are
(A) Ag and Pb
(B) Ag and Cd
(C) Cd and Pb
(D) Cd and Zn
Sol.
A
X  AgNO3 ,
Y  Pb NO3 2
NaCl
PbCl2  Q 
Partly soluble in hot water 
AgCl
P 
AgCl  2NH3   Ag NH3  2  Cl
KI
AgCl  2Na2 S2 O3 
 Na3  Ag  S2 O3 2   2NaCl
PbI2
 Yellow 
*4.
Newman projections P, Q, R and S are shown below:
Which one of the following options represents identical molecules?
(A) P and Q
(B) Q and S
(C) Q and R
(D) R and S
Sol.
(I)
C
HO
HO
Me
Me
Me
Me
Et
Et
Me
Me
Me
Me
Me
HO
Me
Et
Me
2,3,3  trimehyl pen tan 2 ol
Me
P 
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 7
(II)
H
Me
HO
H
H
Me
Me
Et
Me
Et
Me
HO
or
Me C
Me
OH
 Q
3  ethyl,2  methyl pen tan 2ol
OH
Et
Et
OH
Me
Me
H
 Q
Me
Me
Et
H
Et
H
H
Et
H
Et
H
OH
Me
(III)
Me
H
Et
H
Et
Me
H
Et
H
Me
OH
R  Me
Me
3  ethyl, 2  methylpen tan 2ol
(IV)
Et
Et
CHMe2
Me
OH
H
H
H
OH
CHMe2
H
Me
OH
Et
HC
H
H
Me
Me
Me
S
3  ethyl, 2  methyl pentan  3ol
Only (Q & R) are same.
*5.
Which one of the following structures has the IUPAC name 3-ethynyl-2-hydroxy-4-methylhex-3-en-5ynoic acid?
(A)
(B)
(C)
Sol.
(D)
D
OH
H
C
C
3
4
H
C
6
C
5
2 COOH
1
3  ethynyl, 2  hydroxy, 4  methyl hex  3  en  5  ynoic acid
CH3
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 8
*6.
The Fischer projection of D-erythrose is shown below:
D—Erythrose and its isomers are listed as P, Q, R and S in Column-I. Choose the correct relationship of P,
Q, R and S with D-erythrose from Column-II.
(A) P  2, Q  3, R  2, S  2
(C) P  2, Q  1, R  1, S  3
Sol.
(B) P  3, Q  1, R  1, S  2
(D) P  2, Q  3, R  3, S  1
C
Given structure : (2R, 3R)
P   2R,3R  , so, identical. Hence P  2
Q   2S,3R  , so, Diastereomers. Hence Q  1
R   2R,3S  , so, Diastereomers. Hence R  1
S   2S,3S  , so, Enantiomers. Hence S  3
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-1 9




*7.
SECTION-2 (Maximum Marks: 24)
This section contains SIX (06) questions
Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is/are correct
answer(s).
For each question, choose the option(s) corresponding to (all) the correct answer(s).
Answer to each question will be evaluated according to the following making scheme:
Full Marks
: +4 If Only (all) the correct option(s) is/are chosen;
Partial Marks
: +3 If all the four option are correct but ONLY three options are chosen;
Partial Marks
: +2 If three or more option are correct but ONLY two options are chosen, both of which
are correct
Partial Marks
: +1 If two or more option are correct but ONLY one option are chosen, and it is a
correct option
Zero Marks
: 0 If none of the options is chosen (i.e. the question is unanswered);
Negative Marks : 2 In all other cases
In thermodynamics, the P – V work done is given by
w    dVPext
For a system undergoing a particular process, the work done is
a 
 RT
w    dV 
 2
 V b V 
This equation is applicable to a
(A) system that satisfies the van der Waal’s equation of state
(B) process that is reversible and isothermal
(C) process that is reversible and adiabatic
(D) process that is irreversible and at constant pressure
Sol.
A, B, C
Wrev   PdV
For one mole of a real gas,
a 

 P  V 2   V  b   RT


 RT
a 
P

  V  b  V 2 


 RT
a 
So, Wrev    

dV …. (1)
  V  b  V 2 


So, above Equation is valid for all reversible processes and for real gases.
*8.
With respect to the compounds I – V, choose the correct statement(s).
(A) The acidity of compound I is due to delocalization in the conjugate base
(B) The conjugate base of compound IV is aromatic
(C) Compound II becomes more acidic, when it has a NO 2 substituent
(D) The acidity of compounds follows the order I > IV > V > II > III
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 0
Sol.
A, B, C
H
Ph
H
H
Ph
C
CH4
H
Ph
I
H
C
C
H
III
II
 pK a  31.5 
pK a  43 
 pK a  48 
V
IV
 pK a  25 
 pK a  16 
Correct order and acidity = IV > V > I > II > III (Decreasing acidity)
*9.
Sol.
In the reaction scheme shown below, Q, R and S are the major products.
The correct structure of
(A) S is
(B)
Q is
(C)
(D)
S is
R is
B, D
O
Me 3C
Me
O
Me
Me
Me 3C
O


AlCl
3
Friedel Craft.acylation
O
C
Me
Me
Me
i ZnHg /HCl


 ii H3PO4 ,
O
OH
C
Me
i CH3MgBr

ii H3O 
HO
Me
R 
Me
Me 3C
Me
Conc.H SO
2
4



 H2O
Me
O
 Q
Me 3C
Me
C
Me
Me
Me
S 
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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10.
Choose the correct statement(s) among the following:
(A) FeCl4  has tetrahedral geometry

(B) Co  en   NH3  2 Cl2 ] has 2 geometrical isomers
(C) FCl4  has higher spin – only magnetic moment than Co  en   NH3 2 Cl2 ]

(D) The cobalt ion in Co  en   NH3  2 Cl2 ] has sp3 d2 hybridization
Sol.
A, C
(A) FeCl4  : Fe53  , weak ligand, so tetrahedral.

d 

(B) Co  en  NH3 2 Cl2  : Three geometrical isomers.
Cl
NH3
N
N
Cl
en
en
Co
N
NH3
NH3
en
Co
N
Cl
NH3
Cl
NH3
N
Cl
Co
N
NH3
Cl
(C) FeCl4  : N  5.   5  5  2   35 BM


 
Co  en  NH3 2 Cl2  : Co3  d6 , diamagnetic
(D) Co  en NH3 2 Cl2  : d2 sp3 .N  0,   0 & FeCl4   Tetrahedral (sp3)


11.
With respect to hypochlorite, chlorate and perchlorate ions, choose correct statement(s).
(A) The hypochlorite ion is strongest conjugate base
(B) The molecular shape of only chlorate ion is influenced by the lone pair of electrons of Cl
(C) The hypochlorite and chlorate ions disproportionate to give rise to identical set of ions
(D) The hypochlorite ion oxidizes the sulfite ion
Sol.
A, B, D
(A) HClO  HClO3  HClO4 (increasing acidic character)
So, ClO  is the strongest conjugate base.
(B) Only Cl  O  and ClO3 has lone pair of electrons on Cl-atom. But Cl  O  is always linear ion as it is
diatomic, so, no effect of lone pair on shape but in ClO3 due the lone pair on Cl shape becomes
pyramidal.
(C) ClO   Cl  ClO3 
  No same set of ions 
ClO3  Cl  ClO4 
(D) ClO  SO23  
 Cl  SO24
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 2
12.
The cubic unit cell structure of a compound containing cation M and anion X is shown below. When
compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s).
(A) The empirical formula of the compound is MX
(B) The cation M and anion X have different coordination geometries
(C) The ratio of M – X bond length to the cubic unit cell edge length is 0.866
(D) The ratio of the ionic radii of cation M to anion X is 0.414
Sol.
A, C
It can be assumed to be CsCl type solid (bcc)
3a
 0.866
2
Formula of solid = MX
r
0.732    0.999
r
So, r  r 



*13.
Sol.
SECTION-3 (Maximum Marks: 24)
This section contains SIX (06) questions. The answer to each question is a NUMBERICAL VLAUE.
For each question, enter the correct numerical value of the answer using the mouse and the on-screen
virtual numeric keypad in the place designated to enter the answer. If the numerical value has more than
two decimal places, truncate round-off the value to TWO decimal places.
Answer to each question will be evaluated according to the following making scheme:
Full Marks
: +4 If ONLY the correct numerical value is entered;
Zero Marks
: 0 In all other cases.
5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette
using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink
color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?
Exp. No.
Vol. of NaOH (mL))
1
12.5
2
10.5
3
9.0
4
9.0
5
9.0
0.11
Vol. of NaOH used at equivalent point = 9 mL
So, mequivalent of NaOH used up to end point = 9   M  1
5  0.1  2  9  M
1
M   0.11
9
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 3
*14.


Consider the reaction A 
 B at 1000 K. At time t’, the temperature of the system was increased to
2000 K and the system was allowed to reach equilibrium. Throughout this experiment the partial pressure
of A was maintained at 1 bar. Given below is the plot of the partial pressure of B with time. What is the
ratio of standard Gibbs energy of the reaction at 1000 K to that at 2000 K?
Sol.
0.25
10
 10
1
100
K 2 at 2000 K 
 100
1
G1o  RT1 ln10  2.303RT1
K1 at 1000 K 
Go2  2.303RT2 log100  2  2.303RT2
G1o 1 1000 1

  0.25
Go2 2  2000 4
15.
Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K.
Its cell reaction is
1
H2  g   O2  g  H2 O   
2
The work derived from the cell on the consumption of 1.0  103 mol of H2  g is used to compress 1.00
mol of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in
K) of the ideal gas?
The standard reduction potentials for the two half – cells are given below.
O 2  g  4H  aq   4e  2H2 O   , E0  1.23 V,
2H  aq   2e   H2  g , E0  0.00 V.
Use F  96500 C mol1, R  8.314 J mol1K 1.
Sol.
13.32
I.
II.
1
O2  g  2H  2e   H2O    ,
2
H2  g  2H  2e  , Eo  0.00 V
E o  1.23V
1
O2  g  
 H2 O,
Eo  1.23 V
2
Go  2  96500  1.23  1  10 3  0.7  166.173 J
H2  g 
W  166.173 J
Wadiabatic 
nR
T  T 
   1 2 1
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 4
1 8.314
 T2  T1 
5 
 3  1


166.173 2
T 
  13.32
8.314
3
166.173 
*16.
Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of
hydrogen gas in litre (L) produced at 300 K and 1.0 atm pressure, when 5.4 g of aluminium and 50.0 mL
of 5.0 M sulfuric acid are combined for the reaction?
(Use molar mass of aluminium as 27.0 g mol1, R  0.082 atm L mol1K 1 )
Sol.
06.15
LR 
2Al 3H2 SO4  Al2  SO 4 3  3H2  g
5.4
27
0.25
 0.2 mol
VH2 
17.
238
92
0.25 mol
0.25  0.0821 300
 6.1575 litre
1
206
U is known to undergo radioactive decay to form 82
Pb by emitting alpha and beta particles. A rock
238
initially contained 68  10 6 g of 92
U . If the number of alpha particles that it would emit during its
238
206
radioactive decay of 92
U to 82
Pb in three half – lives is Z  1018 , then what is the value of Z?
Sol.
1.20
238
92
206


 
U 82 Pb  8 24 He  6 01
68  10 6
238
 0.286  10 6 mol
So, after 3 half-lives fraction left = 0.125
Fraction of uranium reacted = 0.875
8  68  106
So, number of moles of   particles 
 0.875 mol
238
8  68  106  0.875  6.023  1023
So, number of   particle 
238
 12.046  1017
 1.20  1018
Total moles of uranium =
18.
In the following reaction, compound Q is obtained from compound P via an ionic intermediate.
What is the degree of unsaturation of Q?
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 5
Sol.
18
OCH3
O
Ph
O
Ph
C
+
O
Ph


Conc. H2 SO4

Ph
Ph
Ph
Ph
Degree of unsaturation = 18
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 6
PA R T III: MA T H E MA T I CS
SECTION 1 (Maximum Marks: 18)
 This section contains SIX (06) questions.
 Each question has FOUR options. ONLY ONE of these four options is the correct answer.
 For each question, choose the option corresponding to the correct answer.
 Answer to each question will be a evaluated according to the following marking scheme:
Full Marks
: +3 if ONLY the correct option is chosen;
Zero Marks
: 0 if none of these of the options is chosen (i.e. the question is unanswered);
Negative Marks
: –1 in all other cases.
*1.
Suppose a, b denote the distinct real roots of the quadratic polynomial x2 + 20x – 2020 and suppose c, d
denote the distinct complex roots of the quadratic polynomial x2 – 20x + 2020. Then the value of
ac(a – c) + ad(a – d) + bc(b – c) + bd(b – d)
(A) 0
Sol.
(B) 8000
(C) 8080
(D) 16000
D
Given a + b = –20, ab = –2020, c + d = 20, cd = 2020-09-28
a2c – ac2 + a2d – ad2 + b2c – b2c – bc2 + b2d – bd2
= (a2 + b2)(c + d) – (a + b)(c2 + d2)
= ((a + b)2 – 2ab)(c + d) – (a + b)((c + d)2 – 2cd)
= (400 – 2  2020)(20) – (–20)(400 – 2  2020)
= 20  800 = 16000
2.
If the function f : R  R is defined by f(x) = |x|(x – sin x), then which of the following statements is
TRUE?
Sol.
(A) f is one-one, but NOT onto
(B) f is onto, but NOT one-one
(C) f is BOTH one-one and onto
(D) f is NEITHER one-one NOR onto
C
f(x) is odd, continuous function
 x  x  sin x  , x  0
f  x  
 x  x  sin x  , x  0
for x  0, f(x) = 2x – sin x – x cos x = x(1 – cos x) + (x – sin x)  0
for x < 0, f(x) = –2x + sin x + x cos x = x(cos x – 1) – (x – sin x) > 0 as x < 0
 So f(x) strictly increases in (–, )  f(x) is one-one
x   f(x)  
x  – f(x)  –. So f(x) is onto
3.
Let the function f : R  R and g : R  R be defined by
1  x 1 1 x 
e e
.
2
Then the area of the region in the first quadrant bounded by the curves y = f(x), y = g(x) and x = 0 is
f  x   e x 1  e  x 1 and g  x  
(A)  2  3  
1
e  e1 
2
(B)  2  3  
1
e  e1 
2
(C)  2  3  
1
e  e1 
2
(D)  2  3  
1
e  e 1 
2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 7
Sol.
A
 e1  3 
 , g   
0
, x 1

f  x    x 1 1 x
, x 1
e  e
1
g  x    e x 1  e1 x 
2
f(x) = g(x)
1
e x 1  e1 x   e x 1  e1 x 
2
e x 1  3e1 x
e x 1  3
y = g(x)
x=1
x=
e1  3 . Let x =  be P.O.I
1
Required area =

0

1  x 1
1

e  e1 x  dx    e x 1  e1 x    e x 1  e1 x   dx
2
2


1

1 x 1 1 x 1
1
3
e  e 0   ex 1  e1 x
2
2
2
1
1  0
1
0

1
1

1
=  e  e    e  e   e  3e1   e0  3e0 
2
2
1
1 1
3

 4
= e     3 
2
e 2
3

=
=   3  2 
= 2  3  
*4.
1
1
e  
2
e
1
1
e  
2
e
Let a, b and  be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y2 =
4x, and suppose the ellipse
x2
y2
a
b2

2
 1 passes through the point P. If the tangents to the parabola and the
ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
(A)
Sol.
1
(B)
2
1
2
(C)
1
3
(D)
2
5
A
Given a, b,   R+
As P lies on ellipse
2
a
2

y 2
b2
1
P(, 2)
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 8
b2
2
a
(Slope of tangent for parabola at P) (Slope of tangent at P for ellipse) = –1
So equation of tangent at P to ellipse ss, m 
 
b2
2a2
e  1
5.
Sol.


2
 1 , b2 = 2a2
1
1

2
2
2
1
and ,
3
3
respectively. Suppose  is the number of heads that appear when C1 is tossed twice, independently, and
suppose  is the number of heads that appear when C2 is tossed twice, independently. Then the probability
that the roots of the quadratic polynomial x2 – ax +  are real and equal, is
40
20
1
1
(A)
(B)
(C)
(D)
81
81
2
4
B
2
1
For C1 P  h   , for C2 P  h   , given roots of the equation x2 – x +  = 0 equal real roots
3
3
Let C1 and C2 be two biased coins such that the probabilities of getting head in a single toss are
2 = 4

 = 2 and  = 1
==0
2
2
2
1 2  1  2
16 4
20
2
So required probability =    2C1        =


3 3 3 3
81 81 81
3
6.
Consider all rectangles lying in the region



 x, y   R  R : 0  x  and 0  y  2sin  2x  
2


and having one side on the x-axis. The area of the rectangle which has the maximum perimeter among all
such rectangles, is
(A)
Sol.
3
2
(B) 
(C)

2 3
(D)
 3
2
C
0  y  2 sin 2x, 0  x 

2
(, 2 sin 2)
(, 2 sin 2)
(, 0)
for rectangle 2 sin 2 = 2 sin 2     
(, 0)

2



Perimeter = 2( – ) + 4 sin 2 = 2          4 sin 2
2


= 4 –  + 4 sin 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-2 9
dp
1

 4  8cos2  0 , cos2   ,  
d
2
3
d2p
d 2
 16 sin 2  0 so P is max
 
 2 
So, area of rectangle = ( – )2 sin 2 =     2 sin 

3 6
 3 
=

3

2

6
2
2 3
SECTION 2 (Maximum Marks: 24)
 This section contains SIX (06) questions.
 Each question has FOUR options. ONE OR MORE THAN ONE of these four option(s) is(are) the
correct answer(s).
 For each question, choose the option(s) corresponding to (all) the correct answer(s).
 Answer to each question will be a evaluated according to the following marking scheme:
Full Marks
: +4 if only (all) the correct option(s) is(are) chosen;
Partial Marks
: +3 if all the four options are correct but ONLY three options are chosen;
Partial Marks
: +2 If three or more options are correct but ONLY two options are chosen, both of
which are correct;
Partial Marks
: + 1 If two or more options are correct but ONLY one option is chosen and it is a
correct option;
Zero Marks
: 0 if none of the options is chosen (i.e. the question is unanswered);
Negative Marks
: –2 in all other cases.
7.
Let the function f : R  R be defined by f(x) = x3 – x2 + (x – 1) sin x and let g : R  R be an arbitrary
function. Let fg: R  R be the product function defined by (fg)(x) = f(x)g(x). Then which of the following
statements is/are TRUE?
(A) If g is continuous at x = 1, then fg is differentiable at x = 1
(B) If fg is differentiable at x = 1, then g is continuous at x = 1
(C) If g is differentiable at x = 1, then fg is differentiable at x = 1
(D) If fg is differentiable at x = 1, then g is differentiable at x = 1
Sol.
A, C
f:RR
f(x) = x3 – x2 + (x – 1) sin x
g:RR
(A)
If g is continuous at x = 1 then fg is differentiable
Let h(x) = f(x)g(x)
f 1  h  g 1  h   f 1 g 1
RHD = h(1+) = lim
h0
h
f 1  h  g 1  h   0
= lim
h0
h
f 1  h   f  0 
= lim g(1  h)  lim
h0
h 0
h
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3 0
(B)
= g(1) f(1)
LHD at x = 1
f 1  h  g 1  h   f 1 g 1
= lim
h 0
h
f 1  h   f 1
= lim g 1  h   lim
= g(1) f(1)
h0
h 0
h
So, h(x) is differentiable at x = 1
Given h(x) = f(x)  g(x) is differentiable
f 1  h  g 1  h   f 1 g 1
f 1  h  g 1  h   f 1 g 1
lim
 lim
h0
h 0
h
h
lim g 1  h   f  1  lim g 1  h   f  1
h0
(C)
(D)
h 0
f(1)  0 and g(1) is not define
So, can not comment over continuity and differentiability
Given g(x) is differentiable
So, h(x) = f(x) g(x)
h(x) = f(x) g(x) + g(x) f(x), as g(x) is differentiable
= f(1) g(1) + 0 will exist
Same as for B
lim g 1  h   lim g 1  h 
h0
h 0
can not say about differentiability of the g(x)
8.
Let M be a 3  3 invertible matrix with real entries and let I denote the 3  3 identity matrix. If
M–1 = adj(adj M), then which of the following statements is/are ALWAYS TRUE?
(A) M = I
Sol.
*9.
(B) det M = 1
(C) M2 = I
(D) (adj M)2 = I
B, C, D
M–1 = adj (adj M)
M adj M = |(M)| I
= |M| M
 |M–1| = ||M| M|
1
3
 M M  |M| = 1 ..... (1)
M
adj M
 MM
M
adj M = |M|2 M  adj M = M as |M| = 1
 M (adj M) = M2
 M2 = I
adj M = M as |M| = 1
(adj M)2 = I (as adj M = M)
replace M by adj M
(adj M) adj (adj M) = |adj M| I = |M|2 I
Multiplying by M
M(adj M) adj (adj M) = |M|2M
|M| adj(adj M) = |M|2 M
adj (adj M) = |M| M
Let S be the set of all complex numbers z satisfying |z2 + z + 1| = 1. Then which of the following statements
is/are TRUE?
Sol.
(A) z 
1 1
 for all z  S
2 2
(B) |z|  2 for all z  S
(C) z 
1 1
 for all z  S
2 2
(D) The set S has exactly four elements
B, C
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3 1
|z2 + z + 1| = 1
2
1
3

z     1
2
4

Using  inequality
2
z
1
3
  z2  z  1  1
2
4
z
1
2
2

1
4
1 1

..... (1)
2 2
||z2 + z| – 1|  1
0  |z2 + z|  2  |z2 + z|  2
|z|2 – |z|  2  |z|2 – |z| – 2  0
(|z| – 2)(|z| + 1)  0
|z|  2
z
*10.
Let x, y and z be positive real numbers. Suppose x, y and z are the lengths of the sides of a triangle opposite
to its angles X, Y and Z, respectively. If
X
Z
2y
tan  tan 
,
2
2 xyz
then which of the following statements is/are TRUE?
(A) 2Y = X + Z
Sol.
(B) Y = X + Z
(C) tan
x
x

2 yz
(D) x2 + z2 – y2 = xz
B, C
tan
X
Z
2y
 tan 
2
2 2x  y  z


2y


  = (s – x)(s – z)
s  s  x  s  s  z  2s
 s(s – y) = (s – x)(s – z)  z2 + x2 = y2

 Triangle is right angled, y 
2
Y=X+Z
1
xz
X

x
2
tan 

=
2 s  s  x   x  y  z  y  z  x 
yz



2
2




11.
Let L1 and L2 be the following straight lines.
x 1 y
z 1
x 1 y
z 1
L1 :


and L 2 :


1
1
3
3
1
1
Suppose the straight line
x   y 1 z  
L:


l
m
2
lies in the plane containing L1 and L2, and passes through the point of intersection of L1 and L2. If the line L
bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
(A)  –  = 3
(B) l + m = 2
(C)  –  = 1
(D) l + m = 0
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3 2
Sol.
A, B
DCs of L1 :
DCs of L 2 :
1
11
3
,
,
1
11
,
1
,
3
11
1
11 11 11
1  3  1  1 
3
1
1
Since

0




11  11 
11  11 
11 11 11
2 2
4
So DCs of acute angle bisectors are
,
,
11 11 11
D.rs can be written as 1, 1, –2    m  1
L1 and L2 intersect at (1, 0, 1) which also lies on bisector L. Point on L having y-coordinate equal to 1 is
(2, 1, –1)   = 2,  = –1
12.
Which of the following inequalities is/are TRUE?
1
(A)

x cos xdx 
0
Sol.
1
3
8
(B)

x sin x dx 
0
3
10
1
(C)

0
x 2 cos x dx 
1
2
1
(D)
2
 x sin x dx  9
2
0
A, B, D
1
(A)

0
1

x2 
x cos x dx  x  1 
 dx
2! 


0
3
8

(B)

1

x3 
x sin x dx  x  x 
 dx
3! 


0
3
10

1
(C)

1

x 2 cos x dx  x 2 dx
0
0
1

3
1
(D)

0
1

x3 
x sin x dx  x 2  x 
 dx
3! 

2

0
2

9
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3 3
SECTION 3 (Maximum Marks: 24)
 This section contains SIX (06) questions. The answer to each question is a NUMBERICAL VALUE.
 For each question, enter the correct numerical values of the answer using the mouse and the on-screen
virtual numeric keypad in the place designated to enter the answer. If the numerical value has more than
two decimal places, truncate/round-off the value of TWO decimal places.
 Answer to each question will be a evaluated according to the following marking scheme:
Full Marks
: +4 If ONLY the correct numerical value is entered;
Zero Marks
: 0 in all other cases.
*13.
Let m be the minimum possible value of log3  3y1  3y 2  3 y3  , where y1, y2, y3 are real numbers for
which y1 + y2 + y3 = 9. Let M be the maximum possible value of  log3 x1  log3 x 2  log3 x 3  , where x1,
x2, x3 are positive real numbers for which x1 + x2 + x3 = 9. Then the value of log2  m3   log3 M2  is
_____
Sol.
8
y  y 2  y3
2
1
3y1  3y 2  3y 3
3
3
(Using AM  GM)
 log3  3 y1  3 y 2  3 y 3   log3 3 4
4
m=4
1
x  x2  x 3
Similarly, 1
  x1x 2 x3  3
3
1
log3 3   log3 x1  log3 x 2  log3 x 3 
3
3  log3 x1  log3 x2  log3 x3
M=3
log2 m3  log3 M2  8
*14.
Let a1, a2, a3, ..... be a sequence of positive integers in arithmetic progression with common difference 2.
Also, let b1, b2, b3, ..... be a sequence of positive integers in geometric progression with common ratio 2. If
a1 = b1 = c, then the number of all possible values of c, for which the equality
2(a1 + a2 + ..... + an) = b1 + b2 + ..... + bn
holds for some positive integer n, is _____
Sol.
1
2(a1 + a2 + ..... + an) = b1 + b2 + ..... + bn
 2n  1 
n

 2   2c   n  1 2    c 

2

 2 1 
2n  n  1
c n
2  1  2n
2x – 2 > x 2 – x
if x  8
2x – 1 > 2x
if x > 3
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3 4
2n 1
n8
2n  1
as c 1 values of c are not possible for n  8
possible values of n = 3, 4, 5, 6, 7; only satisfy when n = 3 and c = 12
So, c 
*15.
Let f : [0, 2]  R be the function defined by




f  x    3  sin  2x   sin  x    sin  3x   .
4
4


If ,   [0, 2] are such that {x  [0, 2] : f(x)  0} = [, ], then the value of  –  is _____
Sol.
1
 3  sin 2x  sin  x     sin  3x   

4
4


3 



=  3  sin2x  sin  x    sin  3x 

4
4 



  


=  3  sin2x  3  4 sin2  x    sin  x  
4  
4



  


=  3  sin2x  3  4 sin2  x    sin  x  
4  
4




 sin  x    0
4


   x   0  x   0, 2 
4
5
1
 x
 –=1
4
4
16.
Sol.
   
 
In a triangle PQR, let a  QR, b  RP and c  PQ . If
  



a  c  b 
a
a  3 , b  4 and       ,
c  a  b a  b
 2
then the value of a  b is _____
108
 

 
Equation reduces to  4a  3b   c  7a  b
  
But a  b  c  0
  

 
So,  4a  3b    a  b   7a  b
 
a  b  6
 2
a  b  108
17.
For a polynomial g(x) with real coefficients, let mg denote the number of distinct real roots of g(x).
Suppose S in the set of polynomials with real coefficients defined by
S = {(x2 – 1)2(a0 + a1x + a2x2 + a3x3) : a0, a1, a2, a3  R}.
For a polynomial f, let f and f denote its first and second order derivatives, respectively. Then the
minimum possible value of  mf   m f   , where f  S , is _____
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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J E E(ADV ANC E D)-2 02 0-Paper-1-PC M-3 5
Sol.
5
f = (x2 – 1)2(a0 + a1x + a2x2 + a3x3)
= (x + 1)2(x – 1)2 (a0 + a1x + a2x2 + a3x3)
For min value of mf + mf, a0 + a1x + a2x2 + a3x3 must have only 1 real root which is 1 or –1, then mf = 3 &
mf = 2  Answer = 5
18.
Let e denote the base of the natural logarithm. The value of the real number a for which the right hand limit
lim
x 0
is equal to a nonzero real number, is _____
Sol.
1  x 1/ x  e1
xa
1
1  x 1/ x  1
lim
x 0 
xa
e  lim
x  0
ln1 x 
1
e x 
e
xa
  x  x2

 x

 x 2 1

1 e
 1
1  e 2  1
= lim
 lim 

x 0 e 
x 0 e
xa
xa 
For value of limit to be a non-zero real number a = 1
END OF THE QUESTION PAPER
FIITJEE Lt d., F IIT J E E House, 2 9-A, K a l u S a r a i, S a r v a p r iy a Vi h a r , New D el h i -1 1 0 0 1 6, Ph 4 6 1 0 6 0 0 0, 2 6 5 6 9 4 9 3, F a x 2 6 5 1 3 9 4 2
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